contestada

A man is whirling a 0.25 kg ball on a 1.5 m long string at 3 m/s. Find the centripetal acceleration of this ball.

Question 2 options:

0.5 m/s2

13.5 m/s2

6 m/s2

2 m/s2

Respuesta :

Lanuel

The centripetal acceleration of this ball is equal to 12 [tex]m/s^2[/tex]

Given the following data:

  • Diameter = 1.5 m
  • Speed, V = 3 m/s.
  • Mass = 0.25 kg

Radius = [tex]\frac{Diameter}{2} = \frac{1.5}{2} = 0.75 \;meters[/tex]

To find the centripetal acceleration of this ball:

The acceleration of an object along a circular track is referred to as centripetal acceleration.

Mathematically, the centripetal acceleration of an object is given by the formula:

[tex]A_c = \frac{V^2}{r}[/tex]

Where:

  • Ac is the centripetal acceleration.
  • r is the radius of the circular track.

V is the velocity of an object.

Substituting the given parameters into the formula, we have;

[tex]A_c = \frac{3^2}{0.75}\\\\A_c = \frac{9}{0.75}\\\\A_c = \frac{9}{0.75}[/tex]

Centripetal acceleration = 12 [tex]m/s^2[/tex]

Read more: https://brainly.com/question/6082363

Otras preguntas